I am trying to find the integral of $$\int \tan x \sec^3 x dx$$
$$\int \tan x(1+\tan^2 x)\sec x\, dx$$
This gets me nowhere since I get a $\sec^2 x$ derivative with tan substitution so I try something else.
$$\int \frac {\sin x}{\cos x} \frac{1}{\cos^3x} dx$$
$$\int \frac {\sin x}{\cos^4 x} dx$$
$u = \cos x$ $du = -\sin x$
$$\int \frac {-1}{u^4} du$$
$$-1\int {u^{-4}} du$$
$$-1 \frac{u^{-3}}{3}$$
$$\frac{-1}{3\cos^3 x}$$
This for some reason is wrong.
Best Answer
You missed a $-$ sign in the denominator. $$\int x^{n} \ \text dx = \frac{x^{n+1}}{n+1} + C$$ when $n \neq -1$.
So when $n=-4$ you get $$\int x^{-4} \ \text dx = -\frac{1}{3} \cdot x^{-3} +C$$
There is still an easier method of doing this: $$\int \tan{x}\cdot \sec^{3}(x) \ \text dx = \int \tan{x} \cdot \sec{x} \cdot \sec^{2}(x) \ \text dx = \int t^{2} \ \text dt$$ by putting $t=\sec{x}$.